Chapter 13, Problem 107P

(a)

To determine

The temperature must the ring be heated to fit over the shaft.

Answer to Problem 107P

The temperature must the ring be heated to fit over the shaft is $\underset{\_}{44°\text{C}}$.

Explanation of Solution

Write the expression for change in length

$\Delta L=\alpha L_0\Delta T$ (I)

Here, $\Delta L$ is the change in length, $\alpha$ is the coefficient of linear expansion, $\Delta T$ is the change in temperature

Replace, $\Delta T$ by $T-T_0$, and rearrange the equation (I) to obtain an expression for $T$

$\begin{array}{c}\Delta L=\alpha L_0\left(T-T_0\right)\\ T=\frac{\Delta L}{\alpha L_0}+T_0\end{array}$ (II)

Conclusion:

Substitute $0.00200\text{cm}$ for $\Delta L$, $12\times {10}^{-6}{\text{K}}^{-1}$ for $\alpha$, $7.00000\text{cm}$ for $L_0$, $20.0°\text{C}$ for $T_0$ in equation (II)

$\begin{array}{c}T=\frac{0.00200\text{cm}}{12\times {10}^{-6}{\text{K}}^{-1}\times 7.00000\text{cm}}+20.0°\text{C}\\ =43.8°\text{C}\\ \approx 44°\text{C}\end{array}$

Therefore, the temperature must the ring be heated to fit over the shaft is $\underset{\_}{44°\text{C}}$.

(b)

To determine

The temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again.

Answer to Problem 107P

The temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again $\underset{\_}{-21°\text{C}}$

Explanation of Solution

Consider the diameters of shaft and ring are equal

$L_s=L_b$ (I)

Here, $L_s$ is the diameter of steel ring, and $L_b$ is the diameter of brass shaft

According to the expression for fractional change in length $L_s$ can be written as

$\begin{array}{c}\Delta L_s={\alpha }_s{L}_{s0}\Delta T\\ L_s-L_{s0}={\alpha }_s{L}_{s0}\Delta T\\ L_s=L_{s0}+{\alpha }_s{L}_{s0}\Delta T\end{array}$ (II)

Here, $L_{s0}$ is the diameter at absolute temperature, ${\alpha }_s$ is the linear explanation coefficient od steel ring, and $\Delta T$ is the change in temperature

Similarly in case of $L_b$

$L_b=L_{b0}+{\alpha }_b{L}_{b0}\Delta T$ (III)

Here, $L_{b0}$ is the diameter of shaft at absolute temperature, ${\alpha }_{b0}$ is the linear expansion coefficient of shaft, and $\Delta T$ is the temperature change

Equate equation (II) and (III) replace $\Delta T$ by $T-T_0$ and rearrange to obtain an expression for $T$

$\begin{array}{c}{L}_{s0}+{\alpha }_s{L}_{s0}\Delta T=L_{b0}+{\alpha }_b{L}_{b0}\Delta T\\ \left(T-T_0\right)\left({\alpha }_s{L}_{s0}-{\alpha }_b{L}_{b0}\right)=L_{b0}-L_{s0}\\ T=T_0+\frac{L_{b0}-L_{s0}}{\left({\alpha }_s{L}_{s0}-{\alpha }_b{L}_{b0}\right)}\end{array}$ (IV)

Conclusion:

Substitute $20.0°\text{C}$ for $T_0$, $7.00200\text{cm}$ for $L_{s0}$, $7.00000\text{cm }$ for $L_{b0}$, $12\times {10}^{-6}{\text{K}}^{-1}$ for ${\alpha }_{s0}$, $19\times {10}^{-6}{\text{K}}^{-1}$ for ${\alpha }_{b0}$ in equation (IV)

$\begin{array}{c}T=20.0°\text{C}+\frac{7.00200\text{cm}-7.00000\text{cm }}{\left(12\times {10}^{-6}{\text{K}}^{-1}\times 7.00000\text{cm}-19\times {10}^{-6}{\text{K}}^{-1}\times 7.00200\text{cm}\right)}\\ =-21°\text{C}\end{array}$

Therefore, the temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again $\underset{\_}{-21°\text{C}}$.